I've got a bunch of binary files, each containing an embedded string near the end of the file but at different places (only occurs once in each file). I need to extract the part of the file starting at the location of the string till the end of the file and dump it into a new file.
eg. If the file's contents is "AWREDEDEDEXXXERESSDSDS" and the string of interest is "XXX", then the part of the file I need is "XXXERESSDSDS".
What's the easiest way to do this in bash?
I came up with this solution:
ls -1 *.bin | xargs strings -n4 --radix=d -f | grep "string" | awk '{sub(/:/, ""); print $2 " " $1 " " $1".";}' | xargs -l1 split -b && rm *.aa
ls -1 *.bin Print only the filenames with the extension "bin" in a list format
xargs strings -n4 --radix=d -f List all the strings in the file and their positions and include the filename in the output
grep "string" Print lines containing "string" (it only occurs once in each file)
awk '{sub(/:/, ""); print $2 " " $1 " " $1".";}' Remove the colon after the filename added by strings, and print the position of the string, the filename, and the filename with a period (this line is used as the arguments for the split command
xargs -l1 split -b Execute the split command for each line using the output of awk as the rest of the arguments
rm *.aa Delete the first parts of the split files. "aa" is the default suffix for the part of the split files.
There are probably better/faster/safer ways of doing this but it's fine for my purposes.